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Third-degree, with zeros of -3,-1, and 2 and passes through the point (3,6)

Thirddegree With Zeros Of 31 And 2 And Passes Through The Point 36 class=

Sagot :

Since the polynomial must have zeroes at x=-3, x=-1, x=2, then, we can write it as a combination of the factors (x+3), (x+1), (x-2):

[tex]p(x)=k(x+3)(x+1)(x-2)[/tex]

The constant k will help us to adjust the value of the polynomial when x=3:

[tex]\begin{gathered} p(3)=k(3+3)(3+1)(3-2) \\ =k(6)(4)(1) \\ =24k \end{gathered}[/tex]

Since p(3) must be equal to 6, then:

[tex]\begin{gathered} 24k=6 \\ \Rightarrow k=\frac{6}{24} \\ \Rightarrow k=\frac{1}{4} \end{gathered}[/tex]

Therefore, the following polynomial function has zeroes at -3, -1 and 2, and passes through the point (3,6):

[tex]p(x)=\frac{1}{4}(x+3)(x+1)(x-2)[/tex]

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